Q.(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young's double slit experiment.

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1. CBSE 2026Set 55/1/11 markMCQ
   Q.The shape of the interference fringes in Young's double-slit experiment, when the distance between the slit and the screen is very large as compared to the slit-separation, is nearly (A) straight (B) parabolic (C) circular (D) hyperbolic

   ›Reveal solutionSolution

   When the screen is very far from the slits compared to their separation, points of constant path difference lie on nearly straight lines parallel to the slits, making the fringes straight.

   The key to understanding fringe shape lies in recognizing what an interference fringe actually represents: it is the locus of all points on the screen where the path difference from the two slits is constant.

   In Young's double-slit experiment, we have two coherent sources $S_1$ and $S_2$ separated by distance $d$. A point $P$ on the screen at distance $D$ from the slits will show constructive or destructive interference depending on the path difference $\Delta = |S_2P - S_1P|$.

   For a bright fringe of order $n$, we need $\Delta = n\lambda$. The question is: what is the shape of the curve connecting all points $P$ that satisfy this condition?

   Geometry of path difference

   Consider a point $P$ on the screen at coordinates $(x, y)$ if we place the origin midway between the slits. The two slits are at positions roughly $(0, \pm d/2, 0)$ in 3D space, and the screen is at distance $D$ along the perpendicular.

   The exact path difference is:

   $\Delta = \sqrt{D^2 + (y - d/2)^2 + x^2} - \sqrt{D^2 + (y + d/2)^2 + x^2}$

   This is the general equation for a hyperbola in the $xy$-plane. So strictly speaking, fringes are hyperbolic curves.

   The far-field approximation

   Now comes the crucial condition: $D \gg d$ (screen distance much larger than slit separation).

   When $D$ is very large, we can use the binomial approximation. For the path from $S_2$ to $P$:

   $S_2P = D\sqrt{1 + \frac{(y-d/2)^2 + x^2}{D^2}} \approx D + \frac{(y-d/2)^2 + x^2}{2D}$

   Similarly for $S_1P$. The path difference becomes:

   $\Delta \approx \frac{(y-d/2)^2 - (y+d/2)^2}{2D} = \frac{-2yd}{2D} = -\frac{yd}{D}$

   (The $x^2$ terms cancel out.)

   For constant path difference $\Delta = n\lambda$, we get:

   $y = -\frac{n\lambda D}{d} = \text{constant}$ …

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2. CBSE 2026Set 55/2/11 markMCQ
   Q.Two statements are given – one labelled Assertion (A) and the other Reason (R). Select the correct answer from the codes below: (A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). (C) (A) is true, but (R) is false. (D) (A) is false and (R) is also false. Assertion (A) : Light added to light can produce darkness. Reason (R) : When two coherent light waves interfere, there is darkness at the position of destructive interference.

   ›Reveal solutionSolution

   Interference of coherent light waves can indeed produce darkness at points of destructive interference; both statements are true and the reason correctly explains the assertion. The answer is (A).

   Why light can produce darkness

   The assertion sounds paradoxical at first—how can adding light to light create darkness? The resolution lies in understanding that light is a wave phenomenon, and waves don't simply add arithmetically in intensity. Instead, they superpose according to their phase relationship.

   When two coherent light waves (waves with a constant phase relationship) meet, their electric field amplitudes add vectorially. If they arrive in phase, the amplitudes reinforce (constructive interference, brighter light). If they arrive exactly out of phase—crest meeting trough—the amplitudes cancel (destructive interference), and the resultant intensity becomes zero or near-zero. This is darkness produced by adding light to light.

   The reason statement captures precisely this mechanism: destructive interference between coherent waves creates regions of darkness.

   Examining the statements

   1. Assertion (A): "Light added to light can produce darkness"

      This is experimentally verified in phenomena like Young's double-slit experiment, thin-film interference, and Newton's rings. At certain positions on the screen or observation plane, the intensity drops to zero despite light arriving from two sources. The statement is true.

   2. Reason (R): "When two coherent light waves interfere, there is darkness at the position of destructive interference"

      Destructive interference occurs when the path difference between two coherent waves is an odd multiple of half-wavelengths:

      $\Delta = \left(m + \frac{1}{2}\right)\lambda, \quad m = 0, 1, 2, \ldots$ …

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3. CBSE 2026Set 55/3/11 markMCQ
   Q.In which of the following does total internal reflection NOT occur ? (A) Twinkling of stars (B) Brilliance of diamonds (C) Optical fibre (D) Reflecting prism

   ›Reveal solutionSolution

   Total internal reflection requires light to travel from a denser to a rarer medium at an angle greater than the critical angle. Twinkling of stars involves refraction through atmospheric layers of varying density, not TIR. The answer is (A).

   Total internal reflection is a phenomenon that occurs when light traveling in a denser medium strikes the boundary with a rarer medium at an angle greater than the critical angle. At this point, instead of refracting into the second medium, all the light reflects back into the denser medium. The key conditions are: light must go from higher to lower refractive index, and the angle of incidence must exceed the critical angle $\theta_c = \sin^{-1}(n_2/n_1)$ where $n_1 > n_2$.

   Let's examine each option to see where this principle applies:

   1. Twinkling of stars: Starlight enters Earth's atmosphere and passes through layers of air with slightly different temperatures and densities. As light moves through these layers, it undergoes refraction — bending at each boundary because the refractive index changes gradually. Sometimes light bends toward regions of higher density, sometimes toward lower density, but it continues to propagate through the atmosphere. The random fluctuations in atmospheric density cause the apparent position and brightness of stars to change rapidly, creating the twinkling effect. This is purely refraction, not total internal reflection.

   2. Brilliance of diamonds: A diamond has a very high refractive index ($n \approx 2.42$). When light enters a diamond and strikes the internal faces, it often does so at angles greater than the critical angle (which is quite small, around $24.4°$ for diamond-air interface). The light then undergoes total internal reflection multiple times inside the diamond before emerging, creating the characteristic sparkle and brilliance. …

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4. CBSE 2026Set V11 markMCQ
   Q.For total internal reflection of light :

   (a) light should be travelling from rarer medium to denser medium

   (b) light should be travelling from denser medium to rarer medium

   (c) light should be incident along the normal

   (d) angle of incidence should be equal to $90^\circ$

   ›Reveal solutionSolution

   (b) light should be travelling from denser medium to rarer medium …

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5. CBSE 2026Set DS1 markMCQ
   Q.If critical angle in a medium be $\alpha$, then for total internal reflection, the angle of incidence $\beta$ should be:

   i) $\beta < \alpha$

   ii) $\beta > \alpha$

   iii) $\beta = \alpha$

   iv) $\beta \le \alpha$

   ›Reveal solutionSolution

   TIR occurs only for angles of incidence larger than the critical angle, so $\beta > \alpha$.

   Concept. When light travels from a denser to a rarer medium, at the critical angle $\alpha$ the refracted ray just grazes the surface (angle of refraction $=90^\circ$). If the angle of incidence is increased beyond this critical angle, refraction is no longer possible and the light is completely reflected back into the denser medium — this is total internal reflection.

   …

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6. CBSE 2026Set A1 markMCQ
   Q.The phase difference φ is related to path difference λ by (A) (λ/π)φ (B) (π/λ)φ (C) (λ/2π)φ (D) (2π/λ)φ

   ›Reveal solutionSolution

   Phase difference = (2π/λ) × path difference.

   The fundamental relation between phase difference (Δϕ) and path difference (Δx) is

   $\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x.$

   …

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7. CBSE 2026Set A1 markMCQ
   Q.For destructive interference, the path difference should be equal to (A) nλ (B) (2n+1)λ/2 (C) zero (D) infinity

   ›Reveal solutionSolution

   Destructive interference occurs when the path difference is an odd multiple of λ/2.

   Two waves interfere destructively (cancel) when they arrive exactly out of phase, i.e. a phase difference of $\pi,3\pi,5\pi,\dots$. In terms of path difference this means an odd multiple of half a wavelength:

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.The displacement of water molecules at any instant on the surface of water at nodal lines is ______.

   ›Reveal solutionSolution

   Nodal lines are where two overlapping waves are always exactly out of phase, so their displacements cancel completely at every instant, leaving zero net displacement.

   When two coherent water-wave sources produce overlapping ripples, at points on a nodal line the crest of one wave always coincides with the trough of the other (path difference = odd multiple of half wavelength), so d …

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9. CBSE 2026Set ANNUAL1 mark
   Q.When light passes from a denser medium to a rarer medium, and the angle of incidence equals the critical angle, what is the angle of refraction?

   ›Reveal solutionSolution

   The critical angle is DEFINED as the angle of incidence at which the refracted ray just grazes the interface, making the angle of refraction exactly 90 degrees.

   When light travels from a denser medium towards a rarer medium, it bends away from the normal. As the angle of incidence is increased, the angle of refraction increases faster and eventually reaches 90 degrees (the refracted ray travels right along the boundary between the two media) - the angle of incidence at which this happens is called the critical angle (theta_c). By …

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10. CBSE 2026Set ANNUAL1 mark
    Q.Answer in one word/sentence: What is the phenomena that causes a bubble in water to shine brightly?

    ›Reveal solutionSolution

    Air bubbles in water appear to shine brightly (like silver) due to total internal reflection of light at the water-air interface of the bubble.

    Light travelling in the denser medium (water) strikes the surface of the air bubble (a rarer medium) at angles greater than the critical angle for the water-air interface. Since it cannot refract out, it is totally internal …

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11. CBSE 2025Set 55/5/11 markMCQ
    Q.Two coherent light waves, each having amplitude $a$, superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between: (A) 0 and $2a^{2}$ (B) 0 and $4a^{2}$ (C) $a^{2}$ and $2a^{2}$ (D) $2a^{2}$ and $4a^{2}$

    ›Reveal solutionSolution

    When two coherent waves of equal amplitude $a$ interfere, the resultant amplitude varies from $0$ (destructive) to $2a$ (constructive); since intensity is proportional to the square of amplitude, the intensity range is 0 to $4a^2$.

    The heart of this problem lies in understanding how wave superposition affects intensity. When two coherent waves meet, they don't simply add their intensities — instead, their amplitudes add vectorially, and the resulting intensity depends on the square of this net amplitude.

    For light waves, intensity $I$ is proportional to the square of the amplitude: $I \propto A^2$. If we set the proportionality constant to unity for simplicity (which is standard when comparing relative intensities), then $I = A^2$.

    Now let's trace what happens when two coherent waves, each with amplitude $a$, interfere.

    The amplitude addition principle

    At any point on the screen, the two waves arrive with some phase difference $\delta$ (which depends on the path difference). The resultant amplitude is found by vector addition:

    $A_{\text{net}} = a_1 + a_2$

    where $a_1$ and $a_2$ are the individual wave amplitudes treated as phasors. For two waves of equal amplitude $a$ with phase difference $\delta$:

    $A_{\text{net}} = \sqrt{a^2 + a^2 + 2a \cdot a \cos\delta} = a\sqrt{2(1 + \cos\delta)}$

    Using the identity $1 + \cos\delta = 2\cos^2(\delta/2)$:

    $A_{\text{net}} = 2a\left|\cos\frac{\delta}{2}\right|$

    Finding the intensity extremes

    1. Maximum amplitude (constructive interference): When $\delta = 0, 2\pi, 4\pi, \ldots$ (waves in phase), we have $\cos(\delta/2) = 1$, so:

    $A_{\max} = 2a$

    The maximum intensity is:

    $I_{\max} = A_{\max}^2 = (2a)^2 = 4a^2$ …

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12. CBSE 2025Set 55/6/11 markMCQ
    Q.Two coherent waves, each of intensity $I_0$, produce interference pattern on a screen. The average intensity of light on the screen is: (A) zero (B) $I_0$ (C) $2I_0$ (D) $4I_0$

    ›Reveal solutionSolution

    Interference redistributes light energy across the screen but cannot create or destroy it, so the average intensity equals the sum of the two individual intensities: $I_0 + I_0 = 2I_0$. The correct option is (C).

    When two coherent waves meet, they produce bright and dark fringes. At some points they add constructively (bright), at others destructively (dark). The question asks for the average intensity over the whole screen — not the maximum or minimum at any particular point.

    The key insight is energy conservation. The two sources together deliver a fixed amount of energy to the screen. Interference only redistributes this energy — concentrating it in bright fringes and depleting it in dark ones — it cannot create or destroy energy. So the average over the pattern must equal what the two waves would deliver independently.

    Let's confirm this with the intensity formula.

    1. Write the resultant intensity at a point. For two coherent waves of intensity $I_0$ each, meeting with phase difference $\delta$:

    $I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\,\cos\delta = 2I_0(1 + \cos\delta)$

    This varies from $I_{\text{max}} = 4I_0$ (at $\delta = 0, 2\pi, \ldots$) down to $I_{\text{min}} = 0$ (at $\delta = \pi, 3\pi, \ldots$).

    2. Average over the pattern. As you move across the screen, the phase difference $\delta$ sweeps uniformly through all values from $0$ to $2\pi$:

    $\langle I \rangle = 2I_0\left(1 + \langle\cos\delta\rangle\right)$

    3. Evaluate the average of cosine.

       Over a complete cycle, $\langle\cos\delta\rangle = 0$.

    4. Conclude.

       $\langle I \rangle = 2I_0(1 + 0) = 2I_0$ …

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